3 + 5 + 7 + .... to n term is

n (n - 2)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(n + 1)^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n (n + 2)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

n^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $n (n + 2)$
$3 + 5 + 7.......$ to n terms
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$= \frac{n}{2} [2 \times 3 + (n - 1) 2]$
$= \frac{n}{2} [6 + 2 n - 2]$
$= \frac{n}{2} [2 n + 4]$
$= n (n + 2)$

Suggest Corrections

Similar questions

If 1+ $\frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + . . . .$ to n terms is S. Then, S is equal to

1 + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + . . . .

\frac{n (n + 3)}{4}

\frac{n (n + 2)}{4}

\frac{n (n + 1) (n + 2)}{6}

If the general term of a sequence is given by $\frac{(n^{2} + n)}{n}$ , then the sum of first n terms of the sequence is given by

n

\frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + \dots

If the $n^{t h}$ term of the sequence is given by $\frac{(n^{2} - 1)}{(n + 1)}$ , then the sum of first n terms of the sequence is given by

Join BYJU'S Learning Program